Let the first term of a series be T1 = 6 and its rth term Tr = 3Tr-1 + 6r, r = 2, 3, , n. If the sum of the first n terms of this series is 15 (n2 - 12n + 39 ) (4 6n - 5 3n + 1 ), then n is equal to .
Answer
6
Solution
Divide by 3r: Tr3r = Tr-13r-1 + 2r
Tr3r = T13 + (22 + 23 + + 2r ) = 2 + 2r+1 - 4 = 2r+1 - 2
Tr = 2 6r - 2 3r
Sn = 2 6 (6n - 1 )5 - 2 3 (3n - 1 )2
= 35 (4 6n - 5 3n + 1 )
Compare: n2 - 12n + 39 = 3 (n - 6)2 = 0 n = 6
Answer: 6